Wind Loads: Envelope Procedure Example (Ref. ASCE 7-22)
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- čas přidán 20. 01. 2024
- In this video we explore a numerical example of the Envelope Procedure for computing wind pressures on a building. We also compute the base shear due to wind loads.
CORRECTION: At time stamp 7:45, the reference table for Kd is Table 26.6-1 (There is no Table 26.1-1). Shoutout to my student, Robert, for catching this error!
#civilengineering #structure #structuralengineering #wind #windloads #structuralanalysis1 #velocity #pressure #exposure #ASCE #ASCE7 #envelope #envelopeprocedure #baseshear #shear
thank for this video , it is very helpfull . hope you make more videos like this
Thank you for this video. It is very helpful!!
thanks for your video, it was really helpful
mean roof height = eave height when the slope is < 10 deg I believe right?
Thanks for your comment/question. Based on the definition of the "mean roof height" in ASCE 7, for roof angles less than 10 deg, the mean roof height IS PERMITTED to be taken as the roof eave height. So, this is an option but not a requirement. I took it as the actual mean height (20 ft) in part so I can just read the Kh value directly from Table 26.10-1 as 1.08. I will reply to your other comment regarding that value on your other post.
2 questions:
1) For base shear, shouldn’t you take trib area to the diaphragm, so only half the wall height?
2) for base shear, why aren’t you taking any pressure from the roof? Per section 28.3.6, you need to take 8psf multiplied by the roof area of the building projected onto a vertical plane normal to the wind.
Hi Jessica,
Thank you for your great questions! I split up my response to your question 2 into two parts.
1) For the base shear, think of this as an over-simplified vertical cantilever beam with a uniform load and the base is fixed to the ground. The base shear would be like the horizontal force reaction. All of that wind load on the walls has to go somewhere, and that somewhere is the base. Now, if we were going to investigate or design the roof diaphragm, we would consider pressures on the upper half of the wall height. But in this example, since we were explicitly asked about the base shear, we have to think of the base as an external support that all of the applied loads are transferred to.
2a) The minimum loads specified in Section 28.3.6 (8 psf on the roof and 16 psf on walls) are used when designing structural elements of the main wind for resisting system (MWFRS) and should be applied separately from the actual wind pressures. So, let’s say we are designing a roof beam that is part of the MWFRS. We apply the actual pressures to the building (like we did in the example) and determine the wind effects (W) on that beam (shear, moment, etc). We input these wind effects (W) into the load combinations and design the beam. We then, separately, apply the 16 psf and 8 psf to the building and determine the wind effects again (W) on that beam. We input the effects (W) due to these minimum pressures into the load combinations and check that the beam is still adequate. For the sake of this example, though, I was only intending to demonstrate how to compute base shear due to wind loading using the Envelope Procedure.
2b) Since the roof pressures actually relieve the total horizontal shear due to the windward roof pressure being larger than the leeward wind pressure and both sides of the ridge are experiencing suction, we can neglect the roof wind pressures by Section 28.3.3. i.e. The total horizontal shear is greater than that determined by neglecting the wind forces on the roof. [I probably should have stated that in the video somewhere.] Note, though, that if the windward side of the roof was in compression, we could not neglect the roof pressures since their horizontal components (projected on a vertical plane) would both add to the wall pressures (pointing to the right).
I have made a spreadsheet to calculate Kh,Kz values and came up with a bit different number then Table 26.10-1. Should I not round the calculation and just truncate the number at two places? In this example when h=19.5 ft, Kh = 1.08 and when h=20 ft, Kh = 1.09. I know it is not much but it does change the numbers a little.
Exp D
h no round rounded formula
15 1.03504 1.04 = (2.41)(15/1935)^(2/11.5)
20 1.08814 1.09 = (2.41)(20/1935)^(2/11.5)
25 1.1312 1.13 = (2.41)(25/1935)^(2/11.5)
30 1.16765 1.17 = (2.41)(30/1935)^(2/11.5)
40 1.22755 1.23 = (2.41)(40/1935)^(2/11.5)
50 1.27613 1.28 = (2.41)(50/1935)^(2/11.5)
60 1.31724 1.32 = (2.41)(60/1935)^(2/11.5)
70 1.35303 1.35 = (2.41)(70/1935)^(2/11.5)
80 1.38482 1.38 = (2.41)(80/1935)^(2/11.5)
90 1.41348 1.41 = (2.41)(90/1935)^(2/11.5)
100 1.43962 1.44 = (2.41)(100/1935)^(2/11.5)
120 1.48599 1.49 = (2.41)(120/1935)^(2/11.5)
140 1.52637 1.53 = (2.41)(140/1935)^(2/11.5)
160 1.56223 1.56 = (2.41)(160/1935)^(2/11.5)
180 1.59456 1.59 = (2.41)(180/1935)^(2/11.5)
200 1.62405 1.62 = (2.41)(200/1935)^(2/11.5)
250 1.68831 1.69 = (2.41)(250/1935)^(2/11.5)
Nice! The equations for Kz beneath Table 26.10-1 would be the most correct to use since these capture the nonlinear nature of this relationship. Thus, using 1.09 (actually 1.088 when rounded to 4 sig figs for h = 20 ft) is the most correct. The advantage of the Table 26.10-1, though, is that you can directly read the values that are presented, and you are also permitted to use linear interpolation with the tabular values....which is helpful for quick calcs and when folks prepare for the PE.
Since you created a nice spreadsheet, though, you can continue to use that. If you really want to have some fun with that spreadsheet, put some IF statements for the different z-domains and rig it up to handle all the exposure categories. Awesome stuff!
Why is it that you did not use the Wind directionality factor, Kd, in your calculation to find the velocity pressure, qz?
Good day. This is a good question. This is one of the updates of ASCE 7-22. Equation 26.10-1 of ASCE 7-22 does not include "kd" as we have seen in previous versions of ASCE 7. The directionality factor (kd) is now included in the later chapters (27 - 30) when computing the various pressures "p" and forces "F".
@@mathandengineeringwithdr.a9515 I understand. Thank you for your response. Very helpful video!